A parallel-plate capacitor has plates of area `A` and separation `d` and is charged to a potential difference `V`. The charging battery is then disconnected, and the plates are pulled apart until their separation is `2d`. Derive expression in terms of `A, d` and `V` for
(a) the new potential difference
(b) the initial and final stored energies, `U_i` and `U_f` and
(c) the work required to increase the separation of plates from d to `2d`.

Answer

A, B, D

Explanation

a. `q=C_iV=((epsilon_0A)/d)V`
`V_f=q/C_f=((epsilon_0AV//d))/((epsilon_0A//2d))=2V`
b. `U_i=1/2C_iV^2=1/2((epsilon_0A)/d)V^2`
`U_f=1/2C_fV_f^2=1/2((epsilon_0A)/(2d))(2V)^2`
`=((epsilon_0A)/d)V^2`
c. `W=U_f-U-i=1/2((epsilon_A)/d)V^2`